2. (a) We note that the electric field points leftward at both points. UsingGGFq E=0, and orienting our xaxis rightward (so ˆipoints right in the figure), we find GF= +×−FHGIKJ= −×−−1610406 4101918.#.#CNCiN ichwhich means the magnitude of the force on the proton is 6.4 × 10–18N and its direction ˆ( i)
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Magnetic Field, Electric charge, Fundamental physics concepts, 2Eb, electric field points, §22-2